POJ_3126_Prime Path(数论+BFS)
来源:互联网 发布:java简单的小游戏 编辑:程序博客网 时间:2024/06/15 17:02
Prime Path
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9509 Accepted: 5450
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
31033 81791373 80171033 1033
Sample Output
670
Source
Northwestern Europe 2006
题型:数论+搜索
题意:
输入两个四位数素数a、b,现在对a进行变换,每次只能变换一位,并且每次变换之后也是素数。求将a变到b最少需要多少步。
比如1033->1733->3733->3739->3779->8779->8179,即从1033 → 8179至少需要6步;
分析:
果断BFS,搜索出每一位上应该是什么。先筛出10000以内的素数以便用于判断。然后从给定的数开始,逐个改变然后从给定的数开始,逐个改变该数的不同位数上的数字,如果是素数,则将这个数入队,同时步数加1,如果在队列非空时能找到目标素数,则输出此时的步数,退出搜索,否则输出Impossible。该数的不同位数上的数字,如果是素数,则将这个数入队,同时步数加1,如果在队列非空时能找到目标素数,则输出此时的步数,退出搜索,否则输出Impossible。
代码:
#include<queue>#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int p[12345];int a,b,flag;int pos[12345],mark[12345];int step;void isprime(){ memset(p,1,sizeof(p)); p[0]=0; p[1]=0; for(int i=2;i*i<=12340+1;i++){ if(!p[i]) continue; for(int j=(i<<1);j<=12340;j+=i) p[j]=0; }}void bfs(){ memset(pos,0,sizeof(pos)); memset(mark,0,sizeof(mark)); queue<int> q; while(!q.empty()) q.pop(); q.push(a); mark[a]=1; while(!q.empty()){ if(q.front()==b){ flag=1; printf("%d\n",pos[b]); break; } int tmp=q.front(); q.pop(); for(int i=0;i<=9;i++){ if(p[tmp/10*10+i]&&!mark[tmp/10*10+i]){ //个位 mark[tmp/10*10+i]=1; q.push(tmp/10*10+i); pos[tmp/10*10+i]=pos[tmp]+1; } if(p[tmp/100*100+i*10+tmp%10]&&!mark[tmp/100*100+i*10+tmp%10]){ //十位 mark[tmp/100*100+i*10+tmp%10]=1; q.push(tmp/100*100+i*10+tmp%10); pos[tmp/100*100+i*10+tmp%10]=pos[tmp]+1; } if(p[tmp/1000*1000+i*100+tmp%100]&&!mark[tmp/1000*1000+i*100+tmp%100]){ //百位 mark[tmp/1000*1000+i*100+tmp%100]=1; q.push(tmp/1000*1000+i*100+tmp%100); pos[tmp/1000*1000+i*100+tmp%100]=pos[tmp]+1; } if(p[i*1000+tmp%1000]&&!mark[i*1000+tmp%1000]&& tmp%1000+i*1000>=1000){ //千位,注意不能是0 mark[i*1000+tmp%1000]=1; q.push(i*1000+tmp%1000); pos[i*1000+tmp%1000]=pos[tmp]+1; } } }}int main(){ isprime(); int n; while(~scanf("%d",&n)){ while(n--){ flag=0; scanf("%d%d",&a,&b); bfs(); if(!flag) printf("Impossible\n"); } } return 0;}
- POJ_3126_Prime Path(数论+BFS)
- POJ_3126_Prime Path
- POJ3126 Prime Path【数论】【BFS】
- POJ 3126 / Northwestern Europe 2006 Prime Path (数论&BFS)
- hdu 1104 BFS + 数论
- hdu 1664 BFS + 数论
- 1104 Remainder(数论+bfs)
- hdu1104 Remaindr-----bfs+数论
- hdu1104 BFS + 数论
- 数论 + bfs Remider hdu1104
- pku3126 Prime Path BFS
- 【BFS】Prime Path
- BFS | 3126 | Prime Path
- BFS Prime Path
- [POJ3126]Prime Path+BFS
- Prime Path(bfs)
- B - Prime Path(BFS)
- [toj2648 **BFS**]**Prime Path**
- 混csdn的方法
- sicily1134.积木分发
- AndroidManifest.xml文件解析
- 孙其功陪你学之——OpenGL加载OBJ文件库glm.c和glm.h
- Truck History
- POJ_3126_Prime Path(数论+BFS)
- Mysql 去除 重复项
- Show-time
- sicily1190. Reduced ID Numbers
- NYOJ-242-计算球体积-2013年7月25日20:41:35
- 【求一个数组元素的所有排列组合】
- NYOJ-40-公约数和公倍数-2013年7月25日21:18:15
- Highways
- NSUserDefaults简单用法